Problem: Simplify; express your answer in exponential form. Assume $r\neq 0, t\neq 0$. $\dfrac{{(r^{3}t^{-3})^{-5}}}{{(r^{-5}t)^{-4}}}$
Answer: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(r^{3}t^{-3})^{-5} = (r^{3})^{-5}(t^{-3})^{-5}}$ On the left, we have ${r^{3}}$ to the exponent ${-5}$ . Now ${3 \times -5 = -15}$ , so ${(r^{3})^{-5} = r^{-15}}$ Apply the ideas above to simplify the equation. $\dfrac{{(r^{3}t^{-3})^{-5}}}{{(r^{-5}t)^{-4}}} = \dfrac{{r^{-15}t^{15}}}{{r^{20}t^{-4}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-15}t^{15}}}{{r^{20}t^{-4}}} = \dfrac{{r^{-15}}}{{r^{20}}} \cdot \dfrac{{t^{15}}}{{t^{-4}}} = r^{{-15} - {20}} \cdot t^{{15} - {(-4)}} = r^{-35}t^{19}$